prove that the energy remains constant in case of a freely falling body....

A at point A KE+ PE = mgh + 0= mgh

h at point B KE +PE = 0+1/2 mv^{2 }

= 0+ 1/2 m * 2gh ( v^{2 }= u^{2 }+ 2as0

=mgh u =0 , v^{2} = 2as = 2gh)

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Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.

Let us now prove that the above law holds good in the case of a freely falling body.

Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest.

In this case we have to show that the total energy (potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.

Body of Mass 'm' Placed at a Height 'h'

## At A,

Back to TopPotential energy = mgh

Kinetic energyKinetic energy = 0 [the velocity is zero as the object is initially at rest]

Total energy at A = Potential energy + Kinetic energy

Total energy at A = mgh …(1)

## At B,

Back to TopPotential energy = mgh

= mg(h - x) [Height from the ground is (h - x)]Potential energy = mgh - mgx

Kinetic energyThe body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.

v^{2}- u

^{2}= 2aS

Here, u = 0, a = g and S = x

Kinetic energy = mgx

Total energy at B = Potential energy + Kinetic energyTotal energy at B = mgh …(2)## At C,

Back to TopPotential energy = m x g x 0 (h = 0)

Potential energy = 0Kinetic energy

The distance covered by the body is hv^{2} - u^{2} = 2aS

Kinetic energy

Kinetic energy = mgh

Total energy at C = Potential energy + Kinetic energy= 0 + mgh

Total energy at C = mgh …(3)It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.

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* Energy is neither created nor destroyed .

* Energy can only be transformed or changed from one form of energy to another.

*So in case of a freely falling body as we all know if a body is dropped from a height ,then at that height potential energy of body is maximum that is *mgh and kinetic is minimum or zero

So if a body is dropped then potential energy decrease due to radiating or losing height and kinetic energy increase due to its speed

*so at the moment when it reaches ground , the potential energy is minimum that is zero and kinetic energy is maximum that is ?mv^2.

So in this way energy of a freely falling body is conserved ....

Plzzz cheers up...

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